The measure generated by the Cantor staircase and the intersection of the Cantor set with its translate Suppose that $T$ is the shift $\bmod 1$ of the Cantor set by an irrational number $\alpha\in (0,1)$. Consider the measure $\mu$ on the interval $[0,1]$ generated by the Cantor staircase. I'd like to know whether (if it is known) $\mu(T(C)\cap C)=0$.
 A: Yes, $\mu((C+\alpha)\cap C)=0$ for all irrational $\alpha$ and for most rational $\alpha$. This can be inferred from the description of $(C+\alpha)\cap C$ on page 57 of [DH95]. I offer a translation of their terminology  ("(1,1)-string sequence", etc). Assume $\alpha$ is not a ternary rational (i.e., does not admit a finite ternary expansion). In the ternary expansion of $\alpha$, the following digits are restrictive: 


*

*all digits $1$

*digits $0$ that are preceded by an odd number of digits $1$

*digits $2$ that are preceded by an even number of digits $1$  


Each restrictive digit of $\alpha$ forces the corresponding ternary digit of every element $x\in (C+\alpha)\cap C$ to take a particular value, $0$ or $2$. (As opposed to being free to choose between $0$ or $2$.) Thus, every restrictive digit halves the Cantor measure of the intersection. 
It is easy to see that every number that is not a ternary rational has infinitely many restrictive digits. 
For a more elaborate investigation of the Hausdorff measure and dimension of the intersection, see [PP13].

[DH95] Davis, Gregory J.; Hu, Tian You. 
On the structure of the intersection of two middle third Cantor sets. 
Publ. Mat. 39 (1995), no. 1, 43–60. 
[PP13] Pedersen, Steen; Phillips, Jason D. 
On intersections of Cantor sets: Hausdorff measure.
Opuscula Math. 33 (2013), no. 3, 575–598. 
